Since we can replace \(f\)  with \(f-b_1r\)  and \(h\)  with \(h-b_2\) , it suffices for our aim to prove the case when \(b_1=b_2=0\) .
By The Mean Value Theorem for Integrals, there exists \(\xi _1\in [r_0/4,r_0/2]\)  and \(\xi _2\in [r_0,2r_0]\) , such that
\(\frac{r_0}{4}f(\xi _1)=\int _\frac{r_0}{4}^\frac{r_0}{2}f(t)dt,\,\,\,\,r_0f(\xi _2)=\int _{r_0}^{2r_0}f(t)dt,\) 
which yields that
\(|f(\xi _1)|\le 4ar_0,\,\,\,\,|f(\xi _2)|\le ar_0.\) 
Then
\(\left|\int _{\xi _1}^{\xi _2}(f^{\prime }+h)\right|\le |f(\xi _1)-f(\xi _2)|+\int _{\xi _1}^{\xi _2}|h|\le 5ar_0+\int _{r_0/4}^{2r_0}|h|\le 6ar_0.\) 
Using the Mean Value Theorem for Integrals again, we can find \(\xi \in [\xi _1,\xi _2]\) , such that
\((\xi _2-\xi _1)|f^{\prime }(\xi )+h(\xi )|\le 6a r_0.\) 
Noting that \(r_0/2<\xi _2-\xi _1\) , we complete the proof.
\(\hfill \Box \)
Lemma 5.2 For any sufficiently small \(r\) , we have
\(\left|\int _r^{2r}F_1(t)dt\right|<\alpha (r)r^2, \,\,\,\,\int _r^{2r}|H_1|(t)dt<\alpha (r)r,\) 
and
\(\left|\int _r^{2r}F_2(t)dt+6\omega _3r^2\right|<\alpha (r)r^2, \,\,\,\,\int _r^{2r}|H_2-12\omega _3|(t)dt<\alpha (r)r,\) 
where \(\lim \limits _{r\rightarrow 0}\alpha (r)=0\) , and \(\omega _3=2\pi ^2\)  is the volume of the 3-dimensional sphere.

We have
\(\left|\int _{r}^{2r} F_1(t)\right|& \le & \int _{r}^{2r} \int _{S^3} |R(g)| u^2 \Theta dSdt \\& =& \int _{B_{2r}\setminus B_r}|R(g)| u^2dV_{g_0}\\&\le &\left(\int _{B_{2r} \setminus B_{r}} | R(g) u^2 |^2 dV_{g_0}\right)^{\frac{1}{2}}\left(\int _{B_{2r} \setminus B_{r}} dV_{g_0}\right)^{\frac{1}{2}} \\&\le & C\Vert R\Vert _{L^2(B_{2r},g)}r^2.\) 
and
\(\int _{r}^{2r} |H_1(t)| dt& \le & 2\int _{r}^{2r} \int _{\mathbb {S}^3}|R(g)| u^2 \frac{\partial \log u}{\partial t}\Theta d\mathbb {S}^3dt+ 2\int _{r}^{2r} \int _{\mathbb {S}^3}|R(g)| u^2\frac{\partial \log \Theta (t,\theta )}{\partial t}(\Theta d\mathbb {S}^3dt) \\&\le & C(\Vert R\Vert _{L^2(B_{2r},g)}\Vert \nabla \phi \Vert _{L^2(B_{2r}\setminus B_r,g_0)}+\int _{B_{2r}\setminus B_r}|R(g)| u^2\frac{1}{r}dV_{g_0})\\&\le & C(\Vert R\Vert _{L^2(B_{2r},g)}\Vert \nabla \phi \Vert _{L^2(B_{2r}\setminus B_r,g_0)}+Cr\Vert R(g)\Vert _{L^2(B_{2r},g)}.\) 
By Lemma REF ,
\(\Vert \nabla \phi \Vert _{L^2(B_{2r}\setminus B_r,g_0)}\le Cr\Vert Ru^2+R(g_0)\Vert _{L^2(B_{2r}\setminus B_r,g_0)}\le C(\Vert R\Vert _{L^2(B_{2r}\setminus B_r,g)}r+r^2),\) 
hence
\(\int _r^{2r}|H_1(t)|dt\le C(\Vert R\Vert _{L^2(B_{2r}\setminus B_r,g)}r+r^2).\) 
Next, we discuss \(F_2\) . Since
\(-\Delta _{g_0}\phi -|\nabla _{g_0}\phi |^2=c(n)Ru^2-c(n)R(g_0),\) 
we obtain
\(\int _{r}^{2r} F_2(t) dt& =& \int _{B_{2r}\setminus B_r}\Delta _{g_0}\phi dV_{g_0} \\& =& -\int _{B_{2r}\setminus B_r}|\nabla _{g_0} \phi |^2dV_{g_0}-\int _{B_{2r}\setminus B_r}(R(g) u^2-R(g_0))dV_{g_0}.\) 
Note that
\(\int _{B_{2r}\setminus B_r}|R(g) u^2-R(g_0)|dV_{g_0}\le C(\Vert R\Vert _{L^2(B_{2r}\setminus B_r,g)}r^2+r^4).\) 
To get the estimate of \(\int _{r}^{2r}F_2\) , we only need to show that
\(\lim _{r\rightarrow 0}\frac{1}{r^2}\int _{B_{2r}\setminus B_r}|\nabla _{g_0}\phi |^2dV_{g_0}=6\omega _3.\) 
Assume there exists \(r_k\rightarrow 0\) , such that
\(\lim _{k\rightarrow \infty }\frac{1}{r_k^2}\int _{B_{2r_k}\setminus B_{r_k}}|\nabla _{g_0}\phi |^2dV_{g_0}=\lambda \ne 6\omega _3.\) 
Set \(u_k=c_kr_ku(r_kx)\) , where \(c_k\)  is chosen such that \(\int _{\partial B_1}\log u_k=0\) .
By the arguments in Section 3, \(\log u_k(x)\)  converges to
\(-\log |x|^{-2}\)  weakly in \(W^{2,p}_{loc}(\mathbb {R}^4\setminus \lbrace 0\rbrace )\) . Then, after passing to a subsequence,
\(\int _{B_2\setminus B_1}|\nabla _{g_0(r_kx)}\log u_k|^2\)  converges to \(6\omega _3\) , hence
\(\lim _{k\rightarrow +\infty }\frac{1}{r_k^2}\int _{B_{2r_k}\setminus B_{r_k}}|\nabla _{g_0}\phi |^2dV_{g_0}=\lim _{k\rightarrow +\infty }\int _{B_2\setminus B_1}|\nabla _{g_0(r_kx)}\log u_k|^2dV_{g_0(r_kx)}=6\omega _3.\) 
A contradiction.
Lastly, we calculate \(\int _r^{2r}|H_2-12\omega _3|\) :
\(\int _{r}^{2r} |H_2(t) -12\omega _3|dt\le \int _{B_{2r}\setminus B_r}\left||\nabla _{g_0} \phi |^2\frac{\partial \log \Theta }{\partial r}-\frac{12}{\Theta }\right|dV_{g_0}+C\int _{B_{2r}\setminus B_r}|R(g) u^2-R(g_0)|\frac{1}{r}dV_{g_0}.\) 
The same argument as above shows that
\(\lim _{r\rightarrow 0}\frac{1}{r}\int _{B_{2r}\setminus B_r}\left||\nabla _{g_0} \phi |^2\frac{\partial \log \Theta }{\partial r}-\frac{12}{\Theta }\right|dV_{g_0}=0.\) 
This completes the proof.
\(\hfill \Box \)
We will give several applications here. First, we study the integrability of \(\Delta _{g} R\) :
Lemma 5.3 There exists \(r_k \rightarrow 0\)  such that
\(\int _{B_{\frac{1}{2}}\setminus B_{r_k}}\Delta _g R(g) dV_g\rightarrow \int _{\partial B_\frac{1}{2}}\frac{\partial R}{\partial r}dS_g.\) 
Here \(B_r\)  is a ball with respect to \(g_0\) .

We have
\(\int _{B_{\frac{1}{2}}\setminus B_{r}}\Delta _g R(g) dV_g=\int _{\partial B_\frac{1}{2}}n_{g,\partial B_\frac{1}{2}}(R)dS_g-\int _{\partial B_r}n_{g,\partial B_r}(R)dS_g.\) 
Applying Lemma REF  to \(b_1=b_2=0\)  and \(f=F_1\) , \(h=H_1\)  , we deduce this lemma from (REF ).
\(\hfill \Box \)
We recall some basic properties of \(Q\) -curvatures, c.f. [1]}, [2]}. On a 4 dimensional manifold, the Paneitz operator is defined as follows:
\(P_{g_0}\varphi =\Delta _{g_0}^2\varphi +div_{g_0}\left(\frac{2}{3}R_{g_0}\nabla _{g_0}\varphi -2Ric_{g_0}^{ij}\varphi _i\frac{\partial }{\partial x^j}\right).\) 
The \(Q\) -curvature of \(g\)  satisfies the following equations:
\(Q(g)=-\frac{1}{12}\Delta _gR(g)-\frac{1}{4}|Ric(g)|^2+\frac{1}{12}R^2,\) 
\(P_{g_0}\phi +2Q(g_0)=2Q(g)e^{4\phi }.\) 
For simplicity, we define
\(T(\phi )=\frac{1}{3}R_{g_0}\frac{\partial \phi }{\partial r}-Ric_{g_0}(\nabla _{g_0}\phi ,\frac{\partial }{\partial r}).\) 
Lemma 5.4 
There exists \(r_k\rightarrow 0\) , such that
\(\lim _{k\rightarrow +\infty }\int _{B_\frac{1}{2}\setminus B_{r_k}} Q(g)dV_g=\int _{B_\frac{1}{2}}Q(g_0)dV_{g_0}-4\omega _3+\int _{\partial B_\frac{1}{2}}\left(\frac{1}{2}\frac{\partial \Delta _{g_0}\phi }{\partial r}+T(\phi )\right).\) 

We have
\(\int _{B_\frac{1}{2}\setminus B_r}Q_gdV_g&=&\int _{B_\frac{1}{2}\setminus B_r}Q_{g_0}dV_{g_0}+\frac{1}{2}\int _{B_\frac{1}{2}\setminus B_r}P_{g_0}(\phi )dV_{g_0}\\&=&\int _{B_\frac{1}{2}\setminus B_r}Q_{g_0}dV_{g_0}\\&&+\frac{1}{2}\int _{\partial (B_\frac{1}{2}\setminus B_r)}\left(\frac{\partial \Delta _{g_0}\phi }{\partial r}+\frac{2}{3}R_{g_0}\frac{\partial \phi }{\partial r}-2Ric_{g_0}(\nabla _{g_0}\phi ,\frac{\partial }{\partial r})\right)dS_{g_0}.\) 
First, we check that
\(\int _{\partial B_r}\left(\frac{2}{3}R_{g_0}\frac{\partial \phi }{\partial r}-2Ric_{g_0}(\nabla _{g_0}\phi ,\frac{\partial }{\partial r})\right)dS_{g_0}\rightarrow 0.\) 
By (),
\(\lim _{r\rightarrow 0}\int _{\partial B_r}|\nabla _{g_0}\phi |dS_{g_0}= 0.\) 
Then (REF ) follows.
Then, by applying Lemma REF  to \(f=F_2/2\) , \(h=H_2/2\)  and \((b_1,b_2)=(-2\omega _3,6\omega _3)\) , we deduce from (REF )
that there exists \(r_k\) , such that
\(\int _{\partial B_{r_k}}\frac{\partial \Delta _{g_0}\phi }{\partial r}dS_{g_0}\rightarrow 8\omega _3.\) 
Therefore, we complete the proof.
\(\hfill \Box \)
Next, we discuss the relationship between \(\Vert R\Vert _{L^2}\) 
and \(\Vert Riem\Vert _{L^2}\) :
Lemma 5.5 
We have
\(\int _{B_\frac{1}{2}}|Riem(g)|^2dV_g<+\infty .\) 

It is well-known that
\(Riem(g)=W(g)+\frac{1}{2}(Ric(g)-\frac{1}{6}R(g)g){~\wedge \!\!\!\!\!\bigcirc ~}g,\) 
where \(W\)  is the Weyl tensor and \({~\wedge \!\!\!\!\!\bigcirc ~}\)  is the Kulkarni-Nomizu product. Since \(|W|^2dV_g\)  is conformal invariant, we only need to check
\(Ric(g)\in L^2\)  here.
Recall that
\(Q(g)=-\frac{1}{12}\Delta _gR(g)-\frac{1}{4}|Ric(g)|^2+\frac{1}{12}R^2\) ,
which means that
\(\int _{B_\frac{1}{2}\setminus B_r}|Ric(g)|^2dV_g&=&\frac{1}{3}\int _{B_\frac{1}{2}\setminus B_r}R^2dV_{g}-\frac{1}{3}\int _{B_\frac{1}{2}\setminus B_r}\Delta _gR(g)dV_{g}-4\int _{B_\frac{1}{2}\setminus B_r}Q(g)dV_{g}\\&=&\frac{1}{3}\int _{B_\frac{1}{2}}R^2dV_{g}-4\int _{B_\frac{1}{2}}Q(g_0)dV_{g_0}\\&&-\frac{1}{3}\int _{\partial (B_\frac{1}{2}\setminus B_r)}\frac{\partial R}{\partial r}dS_g-2\int _{\partial (B_\frac{1}{2}\setminus B_r)}(\frac{\partial \Delta _{g_0}\phi }{\partial r}+2T(\phi ))dS_{g_0}\) 
Applying Lemma REF  to \(f=\frac{1}{3}F_1+2F_2\) , \(h=\frac{1}{3}H_1+2H_2\) , \((b_1,b_2)=(-8\omega _3,24\omega _3)\) , we can find \(r_k\rightarrow 0\) , such that
\(\frac{1}{3}\int _{\partial B_{r_k}}\frac{\partial R}{\partial r}dS_g+2\int _{\partial B_{r_k}}(\frac{\partial \Delta _{g_0}\phi }{\partial r}+2T(\phi ))\rightarrow 16\omega _3.\) 
 \(\hfill \Box \)
Lastly, we consider the formula for Pfaffian form:
Lemma 5.6 
We have
\(\int _{B_\frac{1}{2}}Pf(g)&=&-4\omega _3+\int _{B_\frac{1}{2}}Pf(g_0)+\frac{1}{2}\int _{\partial B_\frac{1}{2}} \frac{\partial \Delta _{g_0}\phi }{\partial r}dS_{g_0}\\&&+\frac{1}{12}\int _{\partial B_\frac{1}{2}}u^2\frac{\partial R}{\partial r}dS_{g_0}-\frac{1}{12}\int _{B_\frac{1}{2}}\Delta _{g_0}R(g_0)dV_{g_0}\\&&+\int _{\partial B_\frac{1}{2}}T(\phi )dS_{g_0}.\) 

Recall that (c.f. [2]})
\(Pf(g)=\frac{1}{8}|W(g)|^2+\frac{1}{12}R^2-\frac{1}{4}|Ric(g)|^2,\) 
where \(W\)  is the Weyl tensor. Since
\(\int _{B}|W(g)|^2dV_{g}=\int _B|W(g_0)|^2dV_{g_0}<+\infty ,\) 
\(Pf(g)\)  is integrable. Since
\(\begin{aligned}Pf(g) & = \frac{1}{8} |W(g)|^2 dV_g+ Q(g) dV_g+ \frac{1}{12} \Delta _g R(g) dV_g, \\Pf(g_0) & = \frac{1}{8} |W(g_0)|^2 dV_{g_0}+ Q(g_0) dV_{g_0}+ \frac{1}{12} \Delta _{g_0} R(g_0) dV_{g_0},\end{aligned}\) 
and
\(P_{g_0} \phi + 2Q(g_0) =2 Q(g) e^{4 \phi },\) 
where \(g = u^2 g_0 = e^{2 \phi } g_0\) ,
we have
\(\int _{B_\frac{1}{2}\setminus B_r} Pf(g)& =& \int _{B_\frac{1}{2}\setminus B_r} Pf(g_0)+ \frac{1}{2}\int _{B_\frac{1}{2}\setminus B_r} P_{g_0} \phi dV_{g_0}\\&& + \frac{1}{12} \int _{B_\frac{1}{2}\setminus B_r} \Delta _g R(g) dV_g-\frac{1}{12} \int _{B_\frac{1}{2}\setminus B_r} \Delta _{g_0} R(g_0) dV_{g_0}\\& =&\int _{B_\frac{1}{2}\setminus B_r}Pf(g_0)+ \frac{1}{2}\int _{\partial (B_\frac{1}{2}\setminus B_r)} \frac{\partial \Delta _{g_0}\phi }{\partial r}dS_{g_0}\\&&+\frac{1}{12}\int _{\partial (B_\frac{1}{2}\setminus B_r)}u^2\frac{\partial R}{\partial r}dS_{g_0}-\frac{1}{12}\int _{B_\frac{1}{2}\setminus B_r}\Delta _{g_0}R(g_0)dV_{g_0}\\&&+\frac{1}{2}\int _{\partial (B_\frac{1}{2}\setminus B_r)}\left(R_{g_0}\frac{\partial \phi }{\partial r}-2Ric_{g_0}(\nabla _{g_0}\phi ,\frac{\partial }{\partial r})\right)dS_{g_0}.\) 
Apply Lemma REF  to \(f=\frac{1}{12}F_1+\frac{1}{2}F_2\) ,
and \(h=\frac{1}{12}H_1+\frac{1}{2}H_2\) , \((b_1,b_2)=(-2\omega _3,6\omega _3)\) , which suffices to complete the proof.
\(\hfill \Box \)
Theorem REF , REF  and REF  can be deduced from
Lemma REF , REF  and REF  easily.
